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6 pints of a 20 percent solution of alcohol in water are mixed with 4 14 May 2016, 14:11
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 pints of a 10 percent alcohol in water solution. The percentage alcohol in the new solution is

A. 16
B. 15
C. 14
D. 13
E. 12
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A
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 14 May 2016, 20:02
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Percentage of alcohol in new solution
=( 6*.2 + 4 * .1) / 10
=(1.2 + .4 ) / 10
= 1.6/10
=.16
Percentage of alcohol in new solution = 16 %
Answer A
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 15 May 2016, 02:57
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Problem could be solved fast with logic.

If we have equal volumes of both solutions =5 pints, then percentage alcohol in new solution would be 15% which is average of two solutions = (20+10)/2

As we have more pints and high concentration in solution 1, then new solution would be higher than 15%

only one answer fit

Answer: A
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 15 May 2016, 04:44
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New solution composition based on alcohol weight = [(6*20%) + (4*10%)] / 10 = 16% alcohol by weight in the new solution.
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 15 May 2016, 09:23
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Total solution = 6+4= 10

Total alcohol= 20% of 6= 1.2 + 10% of 4= .4
1.2+.4= 1.6

%= 1.6/10*100= 16

'A' is the answer
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 10 Sep 2020, 19:17
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We can use the concept of weighted averages as the quantities are different.

\(P_{Avg} = \frac{P_1Q_1 \space + \space P_2Q_2}{Q_1 + Q_2}\)

Here Q1 and Q2 are the quantities and P1 and P2 are the percentages.

Substituting the given values, we get \(P_{Avg} = \frac{20*6\space + \space 4*10}{6 + 4} = \frac{120 + 40}{10} = \frac{160}{10} = 16\)%


Option A

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6 pints of a 20 percent solution of alcohol in water are mixed with 4 07 Dec 2024, 07:20
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